The normal to the curve y = ax^1/2 + bx at the point where x = 1 has a gradient 1 and intercepts the y-axis (0,-14). Find a and b
y = a√x + bx
y' = a/(2√x) + b
Since the normal at x=1 has slope 1, the tangent has slope -1. So,
y'(1) = a/2 + b = -1
b = -(a+2)/2
y(1) = a - (a+2)/2 = (a-2)/2
The normal line through (1,(a-2)/2) is thus
y-(a-2)/2 = 1(x-1)
This passes through (0,-14), so
-14 - (a-2)/2 = 1(-1)
a = -24
So, b = 11
check:
y = -24/√x + 11x
y(1) = -13
the normal at (1,-13) is
y+13 = 1(x-1)
y = x-14
This passes through (0,-14)
Well, to find a and b, we need to use some clown math! Let's start by finding the derivative of the curve to find the gradient.
The derivative of y = ax^(1/2) + bx with respect to x is:
dy/dx = (1/2)ax^(-1/2) + b
Now, we can plug in the value x = 1 to find the gradient at that point:
dy/dx = (1/2)a + b
We know that the gradient at x = 1 is 1, so we can set this equal to 1 and solve for a and b:
(1/2)a + b = 1
Now, let's look at the y-intercept. We know that the point (0, -14) lies on the curve, so we can substitute x = 0 and y = -14 into the equation:
-14 = a(0)^(1/2) + b(0)
That simplifies to:
-14 = 0 + 0
Well, that doesn't really give us any information about a or b, does it? It seems like we're missing something here. Maybe we should consult with a non-clown math expert to help us out because I seem to have taken a wrong turn in my calculations!
To find the values of a and b, we need to use the given information that the normal to the curve at x = 1 has a gradient of 1 and intercepts the y-axis at (0, -14).
We'll start by finding the gradient of the curve at x = 1. To do this, we'll differentiate the equation y = ax^(1/2) + bx with respect to x.
Differentiating y = ax^(1/2) + bx:
dy/dx = (1/2)ax^(-1/2) + b
Now, substitute x = 1 into the equation to find the gradient at x = 1:
dy/dx = (1/2)a(1)^(-1/2) + b
dy/dx = (1/2)a + b
Given that the gradient at x = 1 is 1, we have:
(1/2)a + b = 1 ...equation 1
Next, we'll find the equation of the line that represents the normal at x = 1. The equation of a line can be written as y = mx + c, where m is the gradient and c is the y-intercept.
Since the gradient of the normal is 1 (perpendicular to the gradient of the curve), the gradient of the line is -1 (negative reciprocal of 1).
At x = 1, the coordinates on the curve are (1, a(1/2) + b). We'll substitute these coordinates into the equation y = mx + c to find c (the y-intercept):
-14 = -1(1) + c
-14 + 1 = c
c = -13
Therefore, the equation of the line representing the normal at x = 1 is:
y = -x - 13 ...equation 2
Now, we'll find the point where the curve intersects with the line representing the normal at x = 1.
Setting the equations for the curve and the line equal to each other:
ax^(1/2) + bx = -x - 13
ax^(1/2) + bx + x + 13 = 0
Let's now substitute x = 1 into this equation:
a(1^(1/2)) + b(1) + 1 + 13 = 0
a + b + 14 = 0 ...equation 3
From equation 1, we have:
(1/2)a + b = 1
Let's now solve equations 1 and 3 simultaneously. Multiply equation 1 by 2 to eliminate the fraction:
a + 2b = 2 ...equation 4
Subtract equation 4 from equation 3:
(a + b + 14) - (a + 2b) = 0 - 2
a + b + 14 - a - 2b = -2
- b + 14 = -2
- b = -16
b = 16
Now, substitute this value of b into equation 1:
(1/2)a + 16 = 1
1/2a = 1 - 16
1/2a = -15
a = -15 * 2
a = -30
Therefore, the values of a and b are -30 and 16 respectively, yielding the equation of the curve as y = -30x^(1/2) + 16x.
To find the values of a and b, we need to use the information given about the curve and its normal.
Let's start by finding the gradient of the curve at the point where x = 1. To do this, we'll differentiate the equation of the curve with respect to x.
Differentiating y = ax^(1/2) + bx:
dy/dx = (1/2)ax^(-1/2) + b
Now, let's find the gradient of the normal to the curve at x = 1. We know that the gradient of the normal is 1, which means the gradient of the curve (dy/dx) multiplied by the gradient of the normal (-1) should equal -1.
So, let's set up the equation:
(1/2)ax^(-1/2) + b * (-1) = -1
Simplifying the equation:
(1/2)ax^(-1/2) - b = -1
Now, let's substitute x = 1 into this equation since we want to find the values of a and b at the point where x = 1.
(1/2)a(1)^(-1/2) - b = -1
Simplifying further:
(1/2)a - b = -1
We also know that the normal intercepts the y-axis at (0, -14). This means when x = 0, y = -14.
Substituting these coordinates into the equation of the curve:
y = ax^(1/2) + bx
-14 = a(0)^(1/2) + b(0)
-14 = 0 + 0
-14 = 0
Since the left-hand side (-14) does not equal zero, this means that the equation y = ax^(1/2) + bx cannot be correct.
Hence, there seems to be an inconsistency or error in the given information. Please re-check the equation or provide any other relevant information, and I will be happy to assist you further in finding the values of a and b.